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I'm going through this "Elementary Differential Equations and Boundary Value Problems" book, by Boyce and DiPrima, and I have a little trouble understanding a passage in the Exact Equations explanation.

"Let $\psi(x,y) = c $ $(1)$ where c is a constant. Assuming that $(1)$ defines $y$ implicitly as a differentiable function of $x$, we can differentiate $(1)$ with respect to x and obtain $\psi_{x}(x,y) + \psi_{y}(x,y)y' = 0$." (ipsis litteris)

How differentiating with respect to $x$ leads up to the second equation exactly? Thanks for any input :)

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Let $f(x,z):= \psi(x,z)$, $g(x):=(x,y(x))$ an $h=f\circ g$. Then, from the chain rule $$D h(x)= Df(g(x)) D g(x)=(\psi_x(x,y(x)) \, , \, \psi_y(x,(y(x))^\top (1 \, , \, y'(x)) = \psi_x(x,y(x)) + \psi_y(x,y(x))y'(x)$$ Since $f$ is constant, the derivative of $h$ is $0$, that is, we have $$ \psi_x(x,y(x)) + \psi_y(x,y(x))y'(x) = 0.$$

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  • $\begingroup$ Yes, you are right. It should be okay now. $\endgroup$ – Niklas Sep 23 '18 at 14:03

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